# Express cos4x as powers of cosx. ?

May 18, 2018

cos4x=cos2(2x)=color(red)[2cos^2(2x)-1

#### Explanation:

$\cos 2 \left(2 x\right) = {\cos}^{2} \left(2 x\right) - {\sin}^{2} \left(2 x\right)$

$= {\cos}^{2} \left(2 x\right) - 1 + {\cos}^{2} \left(2 x\right) = \textcolor{red}{2 {\cos}^{2} \left(2 x\right) - 1}$

$= 2 \left[\cos 2 x \cdot \cos 2 x\right] - 1 = 2 \left[\left({\cos}^{2} x - {\sin}^{2} x\right) \cdot \left({\cos}^{2} x - {\sin}^{2} x\right)\right] - 1$

$= 2 \left[{\cos}^{4} x - {\sin}^{2} x \cdot {\cos}^{2} x - {\sin}^{2} x \cdot {\cos}^{2} x + {\sin}^{4} x\right] - 1$

$= \left[2 {\cos}^{4} x - 4 {\sin}^{2} x \cdot {\cos}^{2} x + 2 {\sin}^{4} x\right] - 1$

May 18, 2018

$\rightarrow \cos 4 x = 8 {\cos}^{4} x - 8 {\cos}^{2} x + 1$

#### Explanation:

$\rightarrow \cos 4 x$

$= \cos 2 \cdot \left(2 x\right)$

$= {\cos}^{2} \left(2 x\right) - {\sin}^{2} \left(2 x\right)$

$= {\left[{\cos}^{2} x - {\sin}^{2} x\right]}^{2} - {\left[2 \sin x \cdot \cos x\right]}^{2}$

$= {\cos}^{4} x - 2 {\cos}^{2} x \cdot {\sin}^{2} x + {\left({\sin}^{2} x\right)}^{2} - 4 {\sin}^{2} x \cdot {\cos}^{2} x$

$= {\cos}^{4} x - 2 {\cos}^{2} x \left(1 - {\cos}^{2} x\right) + {\left(1 - {\cos}^{2} x\right)}^{2} - 4 \left(1 - {\cos}^{2} x\right) \cdot {\cos}^{2} x$

$= {\cos}^{4} x - 2 {\cos}^{2} x + 2 {\cos}^{4} x + 1 - 2 {\cos}^{2} x + {\cos}^{4} x - 4 {\cos}^{2} x + 4 {\cos}^{4} x$

$= 8 {\cos}^{4} x - 8 {\cos}^{2} x + 1$